Gravitation

Class 11 NCERT Physics

NCERT

1   Answer the following:$\\$ (a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?$\\$ (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?$\\$ (c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

Solution :

(a) No. Gravitational influence of matter on nearby objects cannot be screened by any means. This is because gravitational force unlike electrical forces is independent of the nature of the material medium. Also, it is independent of the status of other objects.$\\$ (b) Yes. If the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).$\\$ (c) Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

2   Choose the correct alternative:$\\$ (a) Acceleration due to gravity increases/decreases with increasing altitude.$\\$ (b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).$\\$ (c) Acceleration due to gravity is independent of mass of the earth/mass of the body.$\\$ (d) The formula $-G Mm(1/r_2-1/r_1)$is more/less accurate than the formula $mg(r_2-r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the earth.

(a) Decreases.$\\$ Explanation : Acceleration due to gravity at depth h is given by the relation:$\\$ $g_h=(1-\dfrac{2h}{r_e})g$$\\ Where, \\ R_e=Radius of the Earth\\ g = Acceleration due to gravity on the surface of the Earth\\ It is clear from the given relation that acceleration due to gravity decreases with an increase in height.\\ (b) Decreases.\\ Explanation: Acceleration due to gravity at depth d is given by the relation:\\ g_d=(1-\dfrac{d}{R_e})g$$\\$ It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.$\\$ (c) Mass of the body.$\\$ Explanation: Acceleration due to gravity of body of mass $m$ is given by the relation:$\\$ $g=\dfrac{GM}{R^2}$$\\ Where,\\ G = Universal gravitational constant\\ M = Mass of the Earth\\ R = Radius of the Earth\\ Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.\\ (d) More.\\ Explanation: Gravitational potential energy of two points r_2 and r_1 distance away from the centre of the Earth is respectively given by:\\ V_{r1}=-\dfrac{GmM}{r_1}\\ V_{r2}=-\dfrac{GmM}{r_2}\\ \\ \\ \\ V(r_1)=\dfrac{GmM}{r_1}\\ V(r_2)=\dfrac{GmM}{r_2}$$\\$ $\therefore \text{ Differencein potential energy,} V=V(r_2)-V(r_1)\\ =-GmM(\dfrac{1}{r_2}-\dfrac{1}{r_1})$$\\ Hence, this formula is more accurate than the formula mg ( r_2 - r_1 ). 3 Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth? Solution : Time taken by the Earth to complete one revolution around the Sun,\\ T_e=1\text{year}$$\\$ Orbital radius of the Earth in its orbit, $R_e = 1 AU$$\\ Time taken by the planet to complete one revolution around the Sun,\\ T_p=\dfrac{1}{2}T_e=\dfrac{1}{2}\text{year}$$\\$ Orbital radius of the planet = $R_p$$\\ From Kepler’s third law of planetary motion, we can write:\\ (\dfrac{R_p}{R_e})^3=(\dfrac{T_p}{T_e})^2\\ \dfrac{R_p}{R_e}=(\dfrac{T_p}{T_e})^{\frac{2}{3}}\\ =(\dfrac{\dfrac{1}{2}}{1})^{\frac{2}{3}}\\ =(o.5)^{\frac{2}{3}} =0.63$$\\$ Hence, the orbital radius of the planet will be $0.63$ times smaller than that of the Earth.

4   Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22 * 10 8 m$ . Show that the mass of Jupiter is about one-thousandth that of the sun.

Orbital period of Io, $I_0,T_{10}=1.769 \text{days}$$\\ =1.769*24*60*60 s$$\\$ Orbital radius of Io,$I_0,R_{10}=4.22*10^8 m$$\\ Satellite I _0 is revolving around the Jupiter\\ Mass of the Jupiter is given by the relation:\\ M_j=\dfrac{4\pi^2R^3_{10}}{GT^2_{10}} ..(i)$$\\$ Where,$\\$ $M_j=\text{Mass of Jupiter}\\ G=\text{Universal gravitational constant }$$\\ Orbital period of Earth,\\ R_e=1AU=1.496*10^{11}m$$\\$ Mass of sun is given as:$\\$ $M_s=\dfrac{4\pi^2R^3_e}{GT^2_e}....(ii)\\ \therefore \dfrac{M_x}{M_j}=\dfrac{4 \pi^2 R^3_e}{GT^2_e}*\dfrac{GT^2_{10}}{4\pi^2R^3_{10}}\\ =\dfrac{R^3_e}{R^3_{10}}*\dfrac{T^2_{10}}{T^2_e}\\ =(\dfrac{1.769*24*60*60}{365.25*24*60*60})^2\\ *(\dfrac{1.496*10^{11}}{4.22*10^8})^3\\ =1045.04\\ \therefore \dfrac{M_s}{M_j}\sim 1000\\ M_s \sim 1000*M_j\\ \therefore \dfrac{M_s}{M_j}=\dfrac{4\pi^2R^3_e}{GT^2_e}*\dfrac{GT^2_{10}}{4 \pi^2R^3_{10}}\\ \dfrac{M_s}{M_j}= \dfrac{R^3_e}{T^2_e}*\dfrac{T^2_{10}}{R^3_{10}}\\ =\dfrac{(1.496*10^{11})^3}{(365.25*24*60*60)^2}\\ *\dfrac{(1.769*24*60*60)^2}{(4.22*10^8)^3}\\ \dfrac{M_s}{M_j}=1045.04\approx 1000\\ M_s=1000*M_j$$\\ Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun. 5 Let us assume that our galaxy consists of 2.5 * 10 ^{11} stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 10^ 5 ly . Solution : Mass of our galaxy Milky Way, M=2.5*10^{11} solar mass\\ Solar mass =Mass of Sun=2.0*10^{36} kg$$\\$ Mass of our galaxy,$M=2.5 *10^{11}*2*10^{36} \\ =5*10^{41} kg $$\\ Diameter of Milky Way, d = 10 ^5 ly$$\\$ Radius of Milky Way, $r =5 * 10^ 4 l y$$\\ 1 ly = 9.46 * 10^{15}m\\ \therefore r=5*10^{4}* 9.46* 10^{15}\\ =4.73* 10^{20}m$$\\$ Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:$\\$ $T=(\dfrac{4\pi^2 r^3}{GM})^{\frac{1}{2}}\\ =(\dfrac{4*(3.14)^2*(4.73)^3*10^{60}}{6.67*10^{-11}*5*10^{41}})^{\frac{1}{2}}\\ =(\dfrac{39.48*105.85*10^{30}}{33.35})^{\frac{1}{2}}\\ =(25.27 *10 ^{30})^{\frac{1}{2}} \\ =1.12 * 10^{16} s\\ 1\text{year}=365*324*60*60 s\\ 1s=\dfrac{1}{365 * 24 * 60 * 60}\text{year}\\ \therefore 1.12 * 10^{16} s\\ =\dfrac{1.12*10^{16}}{365 *24 * 60 * 60}\\ 3.55*10^8\text{years}$

6   Choose the correct alternative:$\\$ (a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.$\\$ (b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.$\\$

Solution :

(a) Kinetic Energy.$\\$ Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-satellite system is a bound system, the total energy of the satellite is negative. Thus, the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.$\\$ (b) Less.$\\$ An orbiting satellite acquires a certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

7   Does the escape speed of a body from the earth depend on$\\$ (a) the mass of the body$\\$ (b) the location from where it is projected$\\$ (c) the direction of projection$\\$ (d) the height of the location from where the body is launched?

Solution :

(a) No$\\$ (b) No$\\$ (c) No$\\$ (d) Yes$\\$ Escape velocity of a body from the Earth is given by the relation:$\\$ $v_{esc}=\sqrt{2gR}\quad ....(i)$$\\$ g = Acceleration due to gravity$\\$ R = Radius of the Earth$\\$ It is clear from equation (i) that escape velocity v esc is independent of the mass of the body and the direction of its projection. However, it depends on gravitational potential at the point from where the body is launched. Since this potential marginally depends on the height of the point, escape velocity also marginally depends on these factors

8   A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed,$\\$ (b) angular speed,$\\$ (c) angular momentum,$\\$ (d) kinetic energy,$\\$ (e) potential energy,$\\$ (f) total energy throughout its orbit?$\\$ Neglect any mass loss of the comet when it comes very close to the Sun.

Solution :

(a)No$\\$ (b)No$\\$ (c)Yes$\\$ (d)No$\\$ (e)No$\\$ (f)Yes$\\$ Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular s peed, kinetic, and potential energy varies from point to point in the orbit.

9   Which of the following symptoms is likely to afflict an astronaut in space$\\$ (a) swollen feet,$\\$ (b) swollen face,$\\$ (c) headache,$\\$ (d) orientational problem?

Solution :

(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.$\\$ (b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.$\\$ (c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.$\\$ (d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space.

10   Choose the correct answer from among the given ones:$\\$ The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see figure )

Solution :

(iii) Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient $(\dfrac{dV}{dr})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational